A substance has density of 2 kg `dm^(-3)` and it crystallizes to fcc lattice with edge length equal to 700 pm . The molar mass of the substance is _______ .

Options

A. `74.50 gm mol^(-1)`
B. `103.30 gm mol^(-1)`
C. `56.02gm mol^(-1)`
D. `65.36gm mol^(-1)`

Answer

To find the molar mass of the substance that crystallizes in a face-centered cubic (FCC) lattice with a given density and edge length, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Density (ρ) = 2 kg/dm³ = 2000 g/dm³ (since 1 kg = 1000 g)
- Edge length (a) = 700 pm = 700 x 10^(-12) m = 7 x 10^(-10) m
- For FCC lattice, the number of atoms per unit cell (Z) = 4
- Avogadro's number (N_A) = 6.022 x 10^(23) mol^(-1)

2. **Convert the Density to Appropriate Units:**
- We already converted the density to grams per cubic decimeter (g/dm³), which is suitable for our calculations.

3. **Use the Formula for Density:**
The formula relating density, molar mass, and unit cell volume is:
\[ \rho = \frac{Z \cdot M}{N_A \cdot V} \]
where:
- ρ = density
- Z = number of atoms per unit cell
- M = molar mass
- N_A = Avogadro's number
- V = volume of the unit cell

4. **Calculate the Volume of the Unit Cell:**
The volume (V) of the cubic unit cell is given by:
\[ V = a^3 \]
Substituting the edge length:
\[ V = (7 \times 10^{-10} \, \text{m})^3 = 3.43 \times 10^{-28} \, \text{m}^3 \]

5. **Convert Volume to dm³:**
Since 1 m³ = 1000 dm³:
\[ V = 3.43 \times 10^{-28} \, \text{m}^3 \times 1000 \, \text{dm}^3/\text{m}^3 = 3.43 \times 10^{-25} \, \text{dm}^3 \]

6. **Rearranging the Density Formula:**
Rearranging the density formula to solve for molar mass (M):
\[ M = \frac{\rho \cdot N_A \cdot V}{Z} \]

7. **Substituting the Values:**
Substitute the known values into the equation:
\[ M = \frac{2000 \, \text{g/dm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1} \cdot 3.43 \times 10^{-25} \, \text{dm}^3}{4} \]

8. **Calculating Molar Mass:**
Calculate the molar mass:
\[ M = \frac{2000 \cdot 6.022 \times 10^{23} \cdot 3.43 \times 10^{-25}}{4} \]
\[ M = \frac{2000 \cdot 6.022 \cdot 3.43}{4} \approx 103.30 \, \text{g/mol} \]

### Final Answer:
The molar mass of the substance is approximately **103.30 g/mol**.